Ellipse

dan03uk · #1 $Old$ April 10th, 2009, 09:19 AM

I am trying to find the area of the segement in the ellipse that is highlighted red. The problem I have is that I anly know the measurments for the rectangel. Which means I don't have values for minor and major axis. On the plus side I do know the quardinates where the rectangle on the ellipse meet. I also know the area of the segment for the circle that has a square of 2 by 2. I was wondering that if you strectched the circle from right into the one on the left would the minor axis stay the same. as the rectangle and square have the same hight.

I really need some help on this matter as its been problem for me for the last 6 days. Any help would be really much appreciated as I need to get my dissertation in on radiation and this ellipse problem is a crucial part.

Prove It · #2 $Old$ April 10th, 2009, 10:20 AM

Quote:

Originally Posted by dan03uk $View Post$

I am trying to find the area of the segement in the ellipse that is highlighted red. The problem I have is that I anly know the measurments for the rectangel. Which means I don't have values for minor and major axis. On the plus side I do know the quardinates where the rectangle on the ellipse meet. I also know the area of the segment for the circle that has a square of 2 by 2. I was wondering that if you strectched the circle from right into the one on the left would the minor axis stay the same. as the rectangle and square have the same hight.

I really need some help on this matter as its been problem for me for the last 6 days. Any help would be really much appreciated as I need to get my dissertation in on radiation and this ellipse problem is a crucial part.

Notice that the ellipse is simply the circle being stretched out in one dimension by a factor of 2.

This means that the area of the ellipse is 2 times the area of the circle.

Thus the red section is twice that of the blue section.

$\textrm{Blue Region } = \frac{1}{4}(\pi r^2 - l^2)$

Do you understand how to get the above formula? It's just the area of the circle, minus the area of the square, all divided by 4.

Therefore $\textrm{Red Region } = 2\times\frac{1}{4}(\pi r^2 - l^2)$

$\textrm{Red Region } = \frac{1}{2}(\pi r^2 - l^2)$ .

The area of the square is easy, because you're given the length.

The area of the circle is NOT so easy, because you're not given the radius. But you can find it.

Notice that the diagonal on the square is actually the diameter of the circle?

So you can find the diameter using Pythagoras, and then halve it.

$d^2 = 2^2 + 2^2$

$d^2 = 4 + 4$

$d^2 = 8$

$d = \sqrt{8} = 2\sqrt{2}$

$r = \frac{1}{2}d$

$r = \frac{1}{2}\times 2\sqrt{2}$

$r = \sqrt{2}$ .

So $\textrm{Red Region } = \frac{1}{2}(\pi r^2 - l^2)$

$\textrm{Red Region } = \frac{1}{2}[\pi \times (\sqrt{2})^2 - 2^2]$

$\textrm{Red Region } = \frac{1}{2}(2\pi - 4)$

$\textrm{Red Region } = (\pi - 2) \textrm{ units}^2$ .

Hope that helped

stapel · #3 $Old$ April 10th, 2009, 10:42 AM

For an explanation of the relationship between the area of the circle and the eillipse, being a "stretched out" circle, try here.